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Electronics => Beginners => Topic started by: Camelkos on June 02, 2020, 01:00:29 pm

Title: the art of electronics : switching the high side of a load
Post by: Camelkos on June 02, 2020, 01:00:29 pm
Hi everybody;

i'm stuck at the circuit attached below, can someone please tell me why the collector current is 4,4 as highlighted thank you all.
Title: Re: the art of electronics : switching the high side of a load
Post by: greenpossum on June 02, 2020, 01:04:29 pm
Upper end of R2 is 14.4V lower end about 0.2V so what is 14.2/3.3k?
Title: Re: the art of electronics : switching the high side of a load
Post by: Wimberleytech on June 02, 2020, 01:09:35 pm
Quote from: Camelkos on June 02, 2020, 01:00:29 pm
Hi everybody;

i'm stuck at the circuit attached below, can someone please tell me why the collector current is 4,4 as highlighted thank you all.
Ignoring Vce(sat)
Title: Re: the art of electronics : switching the high side of a load
Post by: Camelkos on June 02, 2020, 01:25:55 pm
I see now, thank you very much for your help !!
Title: Re: the art of electronics : switching the high side of a load
Post by: Zero999 on June 02, 2020, 02:02:34 pm
I prefer to put the resistor on the emitter side of the NPN transistor.
(https://www.eevblog.com/forum/beginners/control-led-from-teensy-uc/?action=dlattach;attach=573140;image)
Title: Re: the art of electronics : switching the high side of a load
Post by: Camelkos on June 02, 2020, 02:31:41 pm
can you please give more details about how thiss ill operate, thank you !!
Title: Re: the art of electronics : switching the high side of a load
Post by: Zero999 on June 03, 2020, 12:28:30 pm
Quote from: Camelkos on June 02, 2020, 02:31:41 pm
can you please give more details about how thiss ill operate, thank you !!
The voltage across R2 is equal to the input voltage, minus Tr2's base-emitter voltage. I = V/R, so the current through R2 and therefore Tr2's emitter depends on its value and the input voltage. Assuming the hFE is very high, the current through  Tr2's collector, is nearly equal to the emitter current. Most of the current will flow through Tr1's base, turning it on. R1 is just there to speed up the turn-off time.

For example.

Suppose the load current is 100mA, the input voltage is 5V and we want to drive Tr1 with 5mA, to ensure it saturates.

VR2 = 5 - VBE = 5 - 0.7 = 4.3V
IR2 = IB = 5mA = 0.005A
R2 = VR2/IR2 = 4.3/0.005 = 860R. Use the nearest standard value of 820R.
R1 can be 1k, which will lower the base current through Tr1 by around 0.7mA, but that doesn't matter.